The decision rule for rejecting the null hypothesis, considering the t-distribution, is of:
- |t| < 1.9801 -> do not reject the null hypothesis.
- |t| > 1.9801 -> reject the null hypothesis.
<h3>What are the hypothesis tested?</h3>
At the null hypothesis, it is tested if there is not enough evidence to conclude that the mean voltage for these two types of batteries is different, that is, the subtraction of the sample means is of zero, hence:

At the alternative hypothesis, it is tested if there is enough evidence to conclude that the mean voltage for these two types of batteries is different, that is, the subtraction of the sample means different of zero, hence:

We have a two-tailed test, as we are testing if the mean is different of a value.
Considering the significance level of 0.05, with 75 + 46 - 2 = 119 df, the critical value for the test is given as follows:
|t| = 1.9801.
Hence the decision rule is:
- |t| < 1.9801 -> do not reject the null hypothesis.
- |t| > 1.9801 -> reject the null hypothesis.
More can be learned about the t-distribution in the test of an hypothesis at brainly.com/question/13873630
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Factor the equation so...
(r^2-pr) and (p^2q-pqr)
Factor out (r^2-pr) = r(r-p)
Factor out (p^2q-pqr) = pq(p-r)
Add a negative to r(r-p) to make it -r(p-r)
(pq-r)(p-r) is the answer... I'm sorry I can't explain things well, but I tried.
Let

where we assume |r| < 1. Multiplying on both sides by r gives

and subtracting this from
gives

As n → ∞, the exponential term will converge to 0, and the partial sums
will converge to

Now, we're given


We must have |r| < 1 since both sums converge, so


Solving for r by substitution, we have


Recalling the difference of squares identity, we have

We've already confirmed r ≠ 1, so we can simplify this to

It follows that

and so the sum we want is

which doesn't appear to be either of the given answer choices. Are you sure there isn't a typo somewhere?
Set the 2x - 1 equal to 9 then work backwards. 2x - 1 = 9 Nine minus one is eight divided by two is four. X = 4
Answer:
Number of hours= 48 hours
Step-by-step explanation:
Giving the following information:
Hourly rate= $7.25
Total earned= $348
<u>To calculate the number of hours worked, we need to use the following formula:</u>
Total earned= hourly rate*number of hours
number of hours= total earned / hourly rate
number of hours= 348 / 7.25
number of hours= 48 hours